Question

Let \(A=\\{1,2,3,4\\}\). Let the functions \(F, G,\) and \(H\) be given with domain and codomain \(A\) defined as \(F(1)=3, F(2)=2, F(3)=2,\) and \(F(4)=4\) \(G(1)=1, G(2)=3, G(3)=4,\) and \(G(4)=2\) \(H(1)=2, H(2)=4, H(3)=1,\) and \(H(4)=3\) Find the following: (a) \(F \circ G\) (b) \(H \circ F\) (c) \(G \circ H\) (d) \(F \circ G \circ H\)

Short answer

(a) {3, 2, 4, 2}; (b) {1, 4, 4, 3}; (c) {3, 2, 1, 4}; (d) {2, 2, 3, 4}.

Step-by-step solution

Understand Composition

The composition of two functions \(F\) and \(G\), denoted \((F \circ G)(x)\), is defined as \(F(G(x))\). This means you first apply the function \(G\) to \(x\), and then apply the function \(F\) to the result.

Calculate F ◦ G

To find \(F \circ G(x)\), we need to compute \(F(G(x))\) for each \(x \in A\). Perform the calculations as follows:- \((F \circ G)(1) = F(G(1)) = F(1) = 3\)- \((F \circ G)(2) = F(G(2)) = F(3) = 2\)- \((F \circ G)(3) = F(G(3)) = F(4) = 4\)- \((F \circ G)(4) = F(G(4)) = F(2) = 2\)Thus, \(F \circ G\) gives the results: {3, 2, 4, 2}.

Calculate H ◦ F

To find \(H \circ F(x)\), we need to compute \(H(F(x))\) for each \(x \in A\). Perform the calculations as follows:- \((H \circ F)(1) = H(F(1)) = H(3) = 1\)- \((H \circ F)(2) = H(F(2)) = H(2) = 4\)- \((H \circ F)(3) = H(F(3)) = H(2) = 4\)- \((H \circ F)(4) = H(F(4)) = H(4) = 3\)Thus, \(H \circ F\) results in: {1, 4, 4, 3}.

Calculate G ◦ H

To find \(G \circ H(x)\), we need to compute \(G(H(x))\) for each \(x \in A\). Perform the calculations as follows:- \((G \circ H)(1) = G(H(1)) = G(2) = 3\)- \((G \circ H)(2) = G(H(2)) = G(4) = 2\)- \((G \circ H)(3) = G(H(3)) = G(1) = 1\)- \((G \circ H)(4) = G(H(4)) = G(3) = 4\)Thus, \(G \circ H\) results in: {3, 2, 1, 4}.

Calculate F ◦ G ◦ H

To find \(F \circ G \circ H(x)\), we need to compute \(F(G(H(x)))\) for each \(x \in A\). Perform the calculations as follows:- \((F \circ G \circ H)(1) = F(G(H(1))) = F(G(2)) = F(3) = 2\)- \((F \circ G \circ H)(2) = F(G(H(2))) = F(G(4)) = F(2) = 2\)- \((F \circ G \circ H)(3) = F(G(H(3))) = F(G(1)) = F(1) = 3\)- \((F \circ G \circ H)(4) = F(G(H(4))) = F(G(3)) = F(4) = 4\)Thus, \(F \circ G \circ H\) yields: {2, 2, 3, 4}.

Key concepts

Discrete Mathematics

Discrete Mathematics is a branch of mathematics that deals with distinct and separate values. Unlike continuous mathematics, which often deals with real numbers and continuous systems, discrete mathematics focuses on sets like numbers that can be counted individually. It encompasses various topics, including logic, set theory, and combinatorics, which are foundational for computer science and information theory.
Function composition, the focus of the original exercise, is a discrete operation where you apply one function to the result of another function. This operation is critical in developing algorithms and understanding complex systems. Here, we were given specific functions with defined values and asked to find their compositions. This pragmatically shows how discrete mathematics can be applied to solve real-world problems by analyzing well-defined input and output sets.

Domain and Codomain

In functions, understanding the domain and codomain is crucial. The domain of a function is the set of all possible input values, while the codomain is the set of possible output values. Together, they define the structure of a function and its limitations within a given context.
In the exercise, the domain and codomain of functions F, G, and H were all set to the set \(A = \{1, 2, 3, 4\}\). This implies that these functions can only take these discrete values as inputs and produce outputs within the same set. Each function transforms each element of the domain into an element in the codomain. This technical understanding ensures that we evaluate functions correctly and respect their inherent properties and constraints.

• The domain and codomain give us the boundary of operation for functions.
• Understanding these sets helps manage expectations about a function's behaviour.
• In computer science, this concept is pivotal when designing algorithms that process discrete data.

Function Operations

Function operations involve various ways to manipulate and combine functions to create new functions or outcomes. One of the most common operations is function composition, denoted by \( (F \circ G)(x) = F(G(x)) \). This operation is all about doing one function after another.
In the original exercise, several compositions like \( F \circ G \) and \( H \circ F \) were given. To find the composition, you follow a simple yet systematic method: take each input \( x \) from the domain, apply the inner function, and then pass the output through the outer function. This helps in building complex functions from simple ones by combining their operations.

• Function composition is not commutative, meaning \( F \circ G eq G \circ F \) in general.
• It's an essential tool for constructing higher-level mathematical models and understanding layered systems.
• These operations can simplify complex problems by breaking down processes into manageable steps.

By mastering function operations, you open new pathways for solving intricate problems, showing the beauty and power of mathematics in action.